Pumpjack - Revision history

Bilka: What is best is subjective because it depends on what you measure by/what your goal is. The wiki aims to be objective and not prescribe player behaviour. Removed speed prod module section. This may fit better in a forum or reddit post.

2025-05-30T11:56:32Z

What is best is subjective because it depends on what you measure by/what your goal is. The wiki aims to be objective and not prescribe player behaviour. Removed speed prod module section. This may fit better in a forum or reddit post.

← Older revision		Revision as of 11:56, 30 May 2025
Line 30:		Line 30:

	2 x (2 + √4 x 1.5) = 10 fluid per second.		2 x (2 + √4 x 1.5) = 10 fluid per second.

	~~Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.~~
	As an example: be two oil wells, W1 and W2, both with an initial yield of 1 000%. The first one is slotted with two normal [[quality]]{{SA}} [[productivity module 3]]s, the second one with two normal quality [[speed module 3]]s. Below is a table summing up the production of each well (amount produced per cycle is rounded up based on the first few lines of this page).
	~~{\| class="wikitable"~~
	\|-
	~~! Time (minutes) !! Produced (W1) !! Yield (W1) !! Produced (W2) !! Yield (W2)~~
	\|-
	~~\| 0 \|\| 0 \|\| 1000 \|\| 0 \|\| 1000~~
	\|-
	~~\| 1 \|\| 5040 \|\| 1000 \|\| 12000 \|\| 1000~~
	\|-
	~~\| 8 \|\| 40 320 \|\| 999 \|\| 96 000 \|\| 997~~
	\|-
	~~\| 72 \|\| 362 830 \|\| 990 \|\| 855 480 \|\| 972~~
	\|-
	~~\| 715 \|\| 3 440 959 \|\| 900 \|\| 7 392 240 \|\| 714~~
	\|-
	~~\| 2000 \|\| 8 718 394 \|\| 720 \|\| 14 510 400 \|\| 200~~
	\|-
	~~\| 5715 \|\| 17 422 423 \|\| 200 \|\| 23 426 400 \|\| 200~~
	\|}

	How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%. W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.

	Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428 (~74h), when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.

	~~Yet higher yields do not change this figure, and e.g. a 5 000 starting yield has W1 and W2 come closest around 369h of operation, with W2 still 5% ahead in terms of total produced.~~

	The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.

	In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack, until mining productivity is researched, at which point speed is best.

	==Gallery==		==Gallery==

Benjaminikuta at 15:18, 11 April 2025

2025-04-11T15:18:07Z

← Older revision		Revision as of 15:18, 11 April 2025
Line 60:		Line 60:
	The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.		The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.

	In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack.		In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack, until mining productivity is researched, at which point speed is best.

	==Gallery==		==Gallery==

Cardinal: /* Tips */ whitespace

2025-04-02T09:25:08Z

Tips: whitespace

← Older revision		Revision as of 09:25, 2 April 2025
Line 19:		Line 19:
	The formula to determine the modified output is:		The formula to determine the modified output is:

	Output of unmodified pumpjack x (1 + number of modules in pumpjack x average module bonus + transmission strength x number of modules in each beacon x average module bonus)		Output of unmodified pumpjack x (1 + number of modules in pumpjack x average module bonus + transmission strength x number of modules in each beacon x average module bonus)

	Example:		Example:

	For a pumpjack on a depleted resource field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:		For a pumpjack on a depleted resource field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:
	2 x (1 + 2 x 0.5 + ~~SQRT(4)~~ x 1.5 x 2 x 0.5) = 10 fluid per second.
			2 x (1 + 2 x 0.5 + √4 x 1.5 x 2 x 0.5) = 10 fluid per second.

	Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:		Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:
	~~2 x (2 + SQRT(4) x 1.5) = 10 fluid per second.~~

			2 x (2 + √4 x 1.5) = 10 fluid per second.

	Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.		Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.
Line 51:		Line 52:
	\|}		\|}

	How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%.		How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%. W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.

	W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.
	Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428 (~74h), when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.		Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428 (~74h), when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.

	Yet higher yields do not change this figure, and e.g. a 5 000 starting yield has W1 and W2 come closest around 369h of operation, with W2 still 5% ahead in terms of total produced.		Yet higher yields do not change this figure, and e.g. a 5 000 starting yield has W1 and W2 come closest around 369h of operation, with W2 still 5% ahead in terms of total produced.


	The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.		The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.


	In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack.		In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack.

Industrious beaver: /* Tips */

2025-04-01T21:10:59Z

Tips

← Older revision		Revision as of 21:10, 1 April 2025
Line 31:		Line 31:

	Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.		Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.
	As an example: be two oil wells, W1 and W2, both with an initial yield of 1 000%. The first one is slotted with two normal [[quality]]{{SA}} [[productivity module 3]]s, the second one with two normal quality [[speed module 3]]s. Below is a table summing up the production of each well.		As an example: be two oil wells, W1 and W2, both with an initial yield of 1 000%. The first one is slotted with two normal [[quality]]{{SA}} [[productivity module 3]]s, the second one with two normal quality [[speed module 3]]s. Below is a table summing up the production of each well (amount produced per cycle is rounded up based on the first few lines of this page).
	{\| class="wikitable"		{\| class="wikitable"
	\|-		\|-

Industrious beaver: /* Tips */

2025-04-01T21:05:12Z

Tips

← Older revision		Revision as of 21:05, 1 April 2025
Line 54:		Line 54:

	W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.		W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.
	Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428, when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.		Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428 (~74h), when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.
			Yet higher yields do not change this figure, and e.g. a 5 000 starting yield has W1 and W2 come closest around 369h of operation, with W2 still 5% ahead in terms of total produced.


	The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.		The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.

Industrious beaver: /* Tips */

2025-04-01T20:37:36Z

Tips

← Older revision		Revision as of 20:37, 1 April 2025
Line 53:		Line 53:
	How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%.		How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%.

	W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 produces more when the well is depleted.		W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 also produces more when the well is depleted.
	Speed modules are thus better from start to finish, even on a well with high initial yield. The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.		Speed modules are thus better from start to finish, even on a well with high initial yield. While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428, when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.
	While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428, when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.

	~~In conclusion: to maximize production of oil and~~ in the ~~absence~~ of ~~beacons, pumpjacks should be equipped with [[speed module]]s. If at least two beacons are present that are slotted~~ with ~~tier 3 speed~~ modules, ~~some quick calculations seem~~ to ~~make tier~~ 3 ~~productivity modules best~~.		The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.


			In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. However, if at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best for the pumpjack.

	==Gallery==		==Gallery==

Industrious beaver at 14:52, 1 April 2025

2025-04-01T14:52:33Z

← Older revision		Revision as of 14:52, 1 April 2025
Line 15:		Line 15:

	== Tips ==		== Tips ==
	~~If the~~ resource field is depleted (to the minimum of 20%), [[speed module]]s are a good option ~~to raise~~ the ~~pumpjacks~~ output. With two [[speed module 3]]s the output doubles from 2 to 4 fluid per second. With two [[quality\|legendary]]{{SA}} speed module 3s, the output increases by a factor of 3.5 from 2 to 7 fluid per second.		A resource field that has depleted (to the minimum of 20%) can still make use of [[speed module]]s, which are thus a good option for raising the pumpjack's output. With two [[speed module 3]]s, the output doubles from 2 to 4 fluid per second. With two [[quality\|legendary]]{{SA}} speed module 3s, the output increases by a factor of 3.5 from 2 to 7 fluid per second.

	The formula to determine the modified output is:		The formula to determine the modified output is:

	Output of unmodified resource field * (1 + (number of modules in pumpjack * average bonus ~~from module)~~ + ~~(number of beacons * distribution efficiency * (~~number of modules in each beacon * average bonus ~~from module))~~).		Output of unmodified pumpjack x (1 + number of modules in pumpjack x average module bonus + transmission strength x number of modules in each beacon x average module bonus)

			Example:

			For a pumpjack on a depleted resource field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:
			2 x (1 + 2 x 0.5 + SQRT(4) x 1.5 x 2 x 0.5) = 10 fluid per second.

			Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:
			2 x (2 + SQRT(4) x 1.5) = 10 fluid per second.


			Additionally, on an infinite resource well (such as oil) and barring the use of beacons, the best modules to place in a pumpjack are [[speed module]]s. This does not depend on yield.
			As an example: be two oil wells, W1 and W2, both with an initial yield of 1 000%. The first one is slotted with two normal [[quality]]{{SA}} [[productivity module 3]]s, the second one with two normal quality [[speed module 3]]s. Below is a table summing up the production of each well.
	{\| class="wikitable"		{\| class="wikitable"
	~~!colspan="2" \| Example~~
	\|-		\|-
	~~\|For 1 pumpjack on a depleted resource field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:~~		! Time (minutes) !! Produced (W1) !! Yield (W1) !! Produced (W2) !! Yield (W2)
	\|2 * (~~1 +~~ (~~2 * 0.5~~) + (4 * 0.75 * (~~2 * 0.5~~)) ~~) = 10 fluid per second.~~		\|-
			\| 0 \|\| 0 \|\| 1000 \|\| 0 \|\| 1000
			\|-
			\| 1 \|\| 5040 \|\| 1000 \|\| 12000 \|\| 1000
			\|-
			\| 8 \|\| 40 320 \|\| 999 \|\| 96 000 \|\| 997
			\|-
			\| 72 \|\| 362 830 \|\| 990 \|\| 855 480 \|\| 972
			\|-
			\| 715 \|\| 3 440 959 \|\| 900 \|\| 7 392 240 \|\| 714
			\|-
			\| 2000 \|\| 8 718 394 \|\| 720 \|\| 14 510 400 \|\| 200
	\|-		\|-
	\|~~Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:~~		\| 5715 \|\| 17 422 423 \|\| 200 \|\| 23 426 400 \|\| 200
	\|~~4 + 2 * (4 * 0.75) = 10 fluid per second.~~
	\|}		\|}

			How to read (e.g. line 3): after both wells have been active for 8 minutes, W1 has produced 40 320 units of oil and its yield has decreased to 999%; W2 has produced 96 000 units of oil and its yield has decreased to 997%.

			W2 has produced more oil than W1 at every point in time. When W1 finally depletes to the minimum 20% of its original yield, it produces (yield x (1+productivity) x speed) oil per second, so 20 x 1.2 x 0.7 = 16.8 oil per second, while W2 produces 20 x 1 x 2 = 40 oil per second, so W2 produces more when the well is depleted.
			Speed modules are thus better from start to finish, even on a well with high initial yield. The difference is exacerbated by an increase in the quality of the modules: with legendary quality modules, the moment W1 depletes, it has produced 21 778 029 units of oil, while W2 has produced 33 713 400 (55% more). The difference in production from there onward is (W1) 20 x 1.5 x 0.7 = 21 to (W2) 20 x 1 x 3.5 = 70.
			While there is a period when W1 produces more than W2 (because W2 depletes quicker), from ~1475 minutes to ~4428 minutes, W2 still has produced more in total than W1 at every point in this period. The closest they come is near minute 4428, when W1 has produced 19 386 927 and W2 20 337 600. Then the gap increases again in favour of W2.

			In conclusion: to maximize production of oil and in the absence of beacons, pumpjacks should be equipped with [[speed module]]s. If at least two beacons are present that are slotted with tier 3 speed modules, some quick calculations seem to make tier 3 productivity modules best.


	==Gallery==		==Gallery==

Cardinal: Fixed link.

2025-03-28T07:14:30Z

Fixed link.

← Older revision		Revision as of 07:14, 28 March 2025
Line 2:		Line 2:
	{{:Infobox:Pumpjack}}		{{:Infobox:Pumpjack}}

	'''Pumpjacks''' extract [[fluid]]s from resource fields. Each field can be covered by only one pumpjack at a fixed spot. The pumpjack will then output an amount of fluid equal to 10 multiplied by the field's yield per second. For example, a field with 538% yield will generate 54 fluid per second. Without [[speed module]]s, one pumpjack cycle takes one second to complete.		'''Pumpjacks''' extract [[fluid system\|fluid]]s from resource fields. Each field can be covered by only one pumpjack at a fixed spot. The pumpjack will then output an amount of fluid equal to 10 multiplied by the field's yield per second. For example, a field with 538% yield will generate 54 fluid per second. Without [[speed module]]s, one pumpjack cycle takes one second to complete.

	Extracting fluid lowers the field's yield by 1% per 300 pumpjack cycles, to a minimum of 20% of the initial yield or 2 fluid per second, whichever is larger.		Extracting fluid lowers the field's yield by 1% per 300 pumpjack cycles, to a minimum of 20% of the initial yield or 2 fluid per second, whichever is larger.

Alfonse: Rewritten to be more resource-neutral. Also, added info about lithium brine being exhaustible.

2025-03-28T05:47:22Z

Rewritten to be more resource-neutral. Also, added info about lithium brine being exhaustible.

← Older revision		Revision as of 05:47, 28 March 2025
Line 2:		Line 2:
	{{:Infobox:Pumpjack}}		{{:Infobox:Pumpjack}}

	'''Pumpjacks''' extract [[~~Crude oil\|crude oil~~]] from ~~oil~~ fields. Each ~~oil~~ field can be covered by only one pumpjack at a fixed spot. The pumpjack will then output an amount of ~~crude oil per cycle~~ equal to 10 multiplied by the field's yield (e.~~g. 54 crude oil for~~ 538% yield) per second. Without [[speed module]]s, one pumpjack cycle takes one second to complete.		'''Pumpjacks''' extract [[fluid]]s from resource fields. Each field can be covered by only one pumpjack at a fixed spot. The pumpjack will then output an amount of fluid equal to 10 multiplied by the field's yield per second. For example, a field with 538% yield will generate 54 fluid per second. Without [[speed module]]s, one pumpjack cycle takes one second to complete.

	~~In [[Space Age]]{{SA}}~~, the ~~pumpjack can also extract [[sulfuric acid]] from [[Vulcanus]]{{SA}}~~, ~~as well as [[lithium brine]]{{SA}} and [[fluorine]]{{SA}} from [[Aquilo]]{{SA}}~~.		Extracting fluid lowers the field's yield by 1% per 300 pumpjack cycles, to a minimum of 20% of the initial yield or 2 fluid per second, whichever is larger.

	~~Extracting oil lowers the field's yield by 1% per 300 pumpjack cycles to a minimum of 20% of the initial yield or 2 oil per second, whichever is larger.~~ They are limited to a maximum output of 1000 ~~crude oil~~ per cycle, achieved by ~~an oil~~ field with more than 9999% yield. However, such a high yield is rare with standard [[map generator]] settings.		They are limited to a maximum output of 1000 fluid per cycle, achieved by a field with more than 9999% yield. However, such a high yield is rare with standard [[map generator]] settings.

	Pumpjacks ~~cannot~~ be placed on ~~a non-oilfield tile~~. Furthermore, the output pipe location is fixed relative to the pumpjack's orientation.		Pumpjacks can only be placed on fluid resource field tiles. Furthermore, the output pipe location is fixed relative to the pumpjack's orientation.

			In the base game, the only kind of resource field yields [[crude oil]]. In [[Space Age]]{{SA}}, the pumpjack can also extract [[sulfuric acid]] from [[Vulcanus]]{{SA}}, as well as [[fluorine]]{{SA}} and [[lithium brine]]{{SA}} from [[Aquilo]]{{SA}}.

			Lithium brine resource fields are special; unlike the other resource fields, these can run out. These patches do not display a percentage; they instead report the total amount of fluid in the field. They also do not produce less fluid over time. A pumpjack will always produce the same amount of fluid per second on any lithium brine field until it is exhausted: 60 fluid per cycle.

	== Tips ==		== Tips ==

Alfonse: /* Tips */ Linking. Also, pumpjacks can be used for more than oil in SA.

2025-03-13T22:22:13Z

Tips: Linking. Also, pumpjacks can be used for more than oil in SA.

← Older revision		Revision as of 22:22, 13 March 2025
Line 11:		Line 11:

	== Tips ==		== Tips ==
	If ~~oil fields are~~ depleted (to the minimum of 20%) ~~speed~~ [[module]]s are a good option to raise the pumpjacks output. With two speed module 3's the output doubles from 2 to 4 ~~crude oil~~ per second. With two legendary ~~speed module 3s~~{{SA}} the output increases by a factor of 3.5 from 2 to 7 ~~crude oil~~ per second.		If the resource field is depleted (to the minimum of 20%), [[speed module]]s are a good option to raise the pumpjacks output. With two [[speed module 3]]s the output doubles from 2 to 4 fluid per second. With two [[quality\|legendary]]{{SA}} speed module 3s, the output increases by a factor of 3.5 from 2 to 7 fluid per second.

	The formula to determine the modified output is:		The formula to determine the modified output is:

	Output of unmodified ~~oil~~ field * (1 + (number of modules in pumpjack * average bonus from module) + (number of beacons * distribution efficiency * (number of modules in each beacon * average bonus from module))).		Output of unmodified resource field * (1 + (number of modules in pumpjack * average bonus from module) + (number of beacons * distribution efficiency * (number of modules in each beacon * average bonus from module))).

	{\| class="wikitable"		{\| class="wikitable"
	!colspan="2" \| Example		!colspan="2" \| Example
	\|-		\|-
	\|For 1 pumpjack on a depleted ~~oil~~ field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:		\|For 1 pumpjack on a depleted resource field with two level 3 speed modules and 4 beacons with each two level 3 speed modules:
	\|2 * (1 + (2 * 0.5) + (4 * 0.75 * (2 * 0.5)) ) = 10 ~~Crude oil~~ per 1 second.		\|2 * (1 + (2 * 0.5) + (4 * 0.75 * (2 * 0.5)) ) = 10 fluid per second.
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	\|Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:		\|Assuming that there will be always two speed 3 modules equipped in the pumpjack and beacons, the formula for 4 beacons can be shortened:
	\|4 + 2 * (4 * 0.75) = 10 ~~crude oil~~ per second.		\|4 + 2 * (4 * 0.75) = 10 fluid per second.
	\|}		\|}